1 Prioritized h-out of-k Resource Allocation for Mobile Ad Hoc Networks and Distributed Systems...

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Prioritized h-out of-k Resource Allocation for

Mobile Ad Hoc Networks and Distributed Systems

Jehn-Ruey JiangHsuan Chuang University

HsinChu, Taiwan

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OutlineMobile Ad hoc NetworkDistributed SystemPrioritized h-out of-k Mutual ExclusionProposed AlgorithmConcluding Remarks

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Mobile Ad hoc Network MANET (by IETF) No Infrastructure, an anytime, anywhere

network Multi-hop communication Link formations occur when nodes move

within transmission range Link failures occur when nodes move out of

transmission range Frequent and unpredictable topology changes

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Mobile Node A

Mobile Node B

Mobile Node D

Mobile Node C

Mobile Node F

Mobile Node E

A-B: Communicate directlyA-F: Communicate via intermediate nodes B and D or via intermediate nodes E and D

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Distributed SystemA set of stationary, autonomous sitesConnected by wired linksCommunicated by message passing

A collection of independent computers that appear to the users as a single computer (by Andrew Tannenbaun)

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Source: Prof. Harondi

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Merits of Distributed SystemResource Sharing:

data, service, devicesComputation Speedup:

PVM, Grid, MetaComputer 10000 CPUs, each running at 50 MIPS, yields 500000 MIPS → instruction to be executed in 0.002 nsec → equivalent to light distance of 0.6 mm → any processor chip of that size would melt immediately

Fault-toleranceto resist failures or to recover when failures occur

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h-out of-k Mutual Exclusion (1/3)

There are k (k1) identical shared resources.

The node may require to access h (1hk) copies out of the k shared resources at a time.

The total number of resources simultaneously accessed by nodes should not exceed k.

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Application:CSCW (Computer Supported Cooperative Work)

Examples:•Multi-pen E-whiteboard

•Mutual driving group

•Cooperative robotics and nanorobotics

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Application:IPAD (Inter-Personal Awareness Devices)

Examples:•Hummingbirds System

•Generalized Hummingbirds System

•Hocman System

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Prioritized h-out of-k MEUsing HPFS (highest priority first se

rve) instead of FCFS (first come first serve)

Can be applied to systems with real time or prioritized applications, where each node is associated with a deadline or a priority

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MANET and Distributed Primitives (1/2)

General approach to implement distributed primitives ( such as mutual exclusion mechanism)

User Applications

Distributed Primitives

Routing Protocol

Mobile Ad Hoc Network

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Our approach to implement distributed primitives ( such as mutual exclusion mechanism)

User Applications

Distributed Primitives Routing Protocol

Mobile Ad Hoc Network

MANET and Distributed Primitives (2/2)

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Proposed Algorithm Overview

Each node has 3 states: ES, CS, NCS Token-based: to maintains a token-oriented DA

G (directed acyclic graph) where the token holder is the sink

the node holding token to check whether

th to enter the CS (Critical Section)

The counter in TOKEN storing the number of available resources

The number of requested resources

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Data Structure state: Indicates node’s state( ES, CS, or NCS) N: The set of all neighbors. height: A triplet (h1, h2, i). hVector: An array of triplets representing node i's v

iew of height of node j, jN . next: Indicates the location of the token from node

i's viewpoint. tokenHolder: Indicates whether or not node i holds

the token. Q: a queue which contains requests of neighbors.

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An ExampleA 5-node system (i.e., n=5), where

each node is labeled 0,…,4.There are 3 resources (i.e., k=3).The TOKEN is held by node 0

initially.The height values form a DAG

initially.

3height=(0, 2, 3) h=0

10

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height=(0, 1, 1) h=0

height=(0, 2, 4) h=0

height=(0, 1, 2) h=0

height=(0, 0, 0) h=0

next varibale

direct link(from higher height to lower height)

k=3 TOKEN(3)

Q:

Q:

Q:

Q:

Q:

Initially

3height=(0, 2, 3) h=0

10

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height=(0, 1, 1) h=2

height=(0, 2, 4) h=0

height=(0, 1, 2) h=0

height=(0, 0, 0) h=0

next varibale


k=3 TOKEN(3)

Q:

Q:

Q:

Q:

Q:

Node 1 requests 2 resources

(1,2)

3height=(0, 2, 3) h=1

10

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height=(0, 1, 1) h=2

height=(0, 2, 4) h=0

height=(0, 1, 2) h=0

height=(0, 0, 0) h=0

next varibale


k=3 TOKEN(3)

Q:

Q:

Q:

Q:

Q:

Node 3 requests 1 resourceNode 1’s request is sent to node 0

(1,2)

(1,2)(3,1)

3height=(0, 2, 3) h=1

10

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height=(0, 1, 1) h=2

height=(0, 2, 4) h=2

height=(0, 1, 2) h=0

height=(0, 0, 0) h=0

next varibale


k=3 TOKEN(3)

Q:

Q:

Q:

Q:

Q:

Node 3’s request is sent to node 1Node 0 sends token to node 1Node 4 requests 2 resources

(3,1)(1,2)

(3,1)

(4,2)

3height=(0, 2, 3) h=1

10

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height=(0, -1, 1) h=2

height=(0, 2, 4) h=2

height=(0, 1, 2) h=0

height=(0, 0, 0) h=0

next varibale


k=3

Q:

Q:

Q:

Q:

Q:

Node 1 sends token to node 3Node 4’s request is sent to nodes 2, 0, 1

(1,2)(4,2)

(3,1)(1,2)

TOKEN(3)

(4,2)

(4,2)

(4,2)

3height=(0,-2, 3) h=1

10

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height=(0, -1, 1) h=2

height=(0, 2, 4) h=2

height=(0, 1, 2) h=0

height=(0, 0, 0) h=0

next varibale


k=3

Q:

Q:

Q:

Q:

Q:

Node 3 enters the CS

(1,2)(4,2)

(1,2)TOKEN(2)

CS

(4,2)

(4,2)

(4,2)

3height=(0, -2, 3) h=1

10

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height=(0, -3, 1) h=2

height=(0, 2, 4) h=2

height=(0, 1, 2) h=0

height=(0, 0, 0) h=0

next varibale


k=3

Q:

Q:

Q:

Q:

Q:

Node 3 sends token to node 1

(1,2)(4,2)

TOKEN(2)

CS

(4,2)

(4,2)

(4,2)

3height=(0, -2, 3) h=1

10

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height=(0, -3, 1) h=2

height=(0, 2, 4) h=2

height=(0, 1, 2) h=0

height=(0, 0, 0) h=0

next varibale


k=3

Q:

Q:

Q:

Q:

Q:


TOKEN(0)

CS

CS

(4,2)

(4,2)

(4,2)

(4,2)

3height=(0, -2, 3) h=1

10

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height=(0, -3, 1) h=2

height=(0, -6, 4) h=2

height=(0, -5, 2) h=0

height=(0, -4, 0) h=0

next varibale


k=3

Q:

Q:

Q:

Q:

Q:

Node 1 sends token to nodes 0, 2, 4

TOKEN(0)

CS

CS

(4,2)

3height=(0, -2, 3) h=0

10

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height=(0, -3, 1) h=2

height=(0, -6, 4) h=2

height=(0, -5, 2) h=0

height=(0, -4, 0) h=0

next varibale


k=3

Q:

Q:

Q:

Q:

Q:

Node 3 leaves the CSNode 3 sends release msg to node 4

TOKEN(1)

CS

(4,2)

3height=(0, -2, 3) h=0

10

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height=(0, -3, 1) h=0

height=(0, -6, 4) h=2

height=(0, -5, 2) h=0

height=(0, -4, 0) h=0

next varibale


k=3

Q:

Q:

Q:

Q:

Q:

Node 1 leaves the CSNode 1 sends release msg to node 4 via node 2

TOKEN(3)

(4,2)

3height=(0, -2, 3) h=0

10

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height=(0, -3, 1) h=0

height=(0, -6, 4) h=2

height=(0, -5, 2) h=0

height=(0, -4, 0) h=0

next varibale


k=3

Q:

Q:

Q:

Q:

Q:


TOKEN(1)

CS

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(cont. 1)

3(0, 2, 3) h=0

10

24

1

(0, 1, 1)h=1

(0, 2, 4) h=0

(0, 1, 2) h=0

1

(0, 0, 0) h=0

k=3

TOKEN(3)

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(cont. 2)

3(0, 2, 3) h=0

10

24

1

(0, 1, 1)h=1

4

(0, 2, 4) h=1

4

(0, 1, 2) h=0

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(0, 0, 0) h=0

k=3

TOKEN(3)

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(cont. 3)

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(0, 2, 3) h=2

10

24

13

(0, 1, 1)h=1

4

(0, 2, 4) h=1

4

(0, 1, 2) h=0

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(0, 0, 0) h=0

k=3

TOKEN(3)

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(cont. 4)

3(0, 2, 3) h=2

10

2

13

(0, 1, 1)h=1

4

(0, 2, 4) h=1

4

(0, 1, 2) h=0

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(0, 0, 0) h=0

k=3

TOKEN(3)

4

3

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(cont. 5)

334

(0, 2, 3) h=2

10

2

13

(0, 1, 1)h=1

4

(0, 2, 4) h=1

4

(0, 1, 2) h=0

12

(0, 0, 0) h=0

k=3

TOKEN(3)

4

=

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(cont. 6)

334

(0, 2, 3) h=2

10

2

13

(0, 1, 1)h=1

4

(0, 2, 4) h=1

(0, 1, 2) h=0

12

(0, 0, 0) h=0

k=3

TOKEN(3)

4

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(cont. 7)

334

(0, 2, 3) h=2

10

2

130

(0, -1, 1)h=1

4

(0, 2, 4) h=1

(0, 1, 2) h=0

12

(0, 0, 0) h=0

k=3 TOKEN(3)

4

=

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(cont. 8)

334

(0, 2, 3) h=2

10

2

130

(0, -1, 1)h=1

4

(0, 2, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3 TOKEN(3)

4

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(cont. 9)

334

(0, 2, 3) h=2

10

2

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(0, -1, 1)h=1

4

(0, 2, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3 TOKEN(2)

4

CS

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(cont. 10)

3341

(0, -2, 3) h=2

10

2

30

(0, -1, 1)h=1

4

(0, 2, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(2)

4

CS

=

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(cont. 11)

3341

(0, -2, 3) h=2

10

2

0

(0, -1, 1)h=1

4

(0, 2, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(2)

4

CS

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(cont. 12)

341

(0, -2, 3) h=2

10

2

0

(0, -1, 1)h=1

4

(0, 2, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(0)

4

CS

CS

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(cont. 13)

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(0, -2, 3) h=2

10

2

0

(0, -1, 1)h=1

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(0, -3, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(0)4

CS

CS

node 4 can’t enter CSsince h(=1) > k (=0)

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(cont. 14)

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(0, -2, 3) h=2

10

2

0

(0, -1, 1)h=0

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(0, -3, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(0)4

CS

RELEASE(1)

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(cont. 15)

31

(0, -2, 3) h=2

10

2

0

(0, -1, 1)h=0

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(0, -3, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(0)4

CS

RELEASE(1)

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(cont. 16)

31

(0, -2, 3) h=2

10

2

0

(0, -1, 1)h=0

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(0, -3, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(1)4

CS

RELEASE(1)

node 4 can enter CSsince h(=1) k (=1)

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(cont. 17)

31

(0, -2, 3) h=2

10

2

0

(0, -1, 1)h=0

43

(0, -3, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(0)4

CS

CS=

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(cont. 18)

31

(0, -2, 3) h=2

10

2

0

(0, -1, 1)h=0

3

(0, -3, 4) h=1

(0, 1, 2) h=0

2

(0, 0, 0) h=0

k=3

TOKEN(0)4

CS

CS

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Concluding Remarks Initially, there is a token holder node i with

height=(0, 0, i). All nodes will eventually form a token-oriented DAG.

The algorithm is sensitive to link forming and link breaking.

If h is always chosen to be 1 k-Mutual Exclusion

If h and k are both chosen to be 1 Mutual Exclusion

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Concluding Remarks (cont.)Restrictions No node failure is allowed (It can be loosed to

be that the token holder and nodes in CS cannot fail.)

The field of height may be unbounded (the breaking and forming of links may help readjust the height value) .

No partition is allowed.

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Concluding Remarks (cont.)Future Work Compare to the Mutual Exclusion Algorithm

[Walter et. al. 1998] and the k-Mutual Exclusion Algorithm [Walter et. al. 2001].

Develop prioritized h-out of-k mutual exclusion algorithm.

Make use of the concept of self-stabilizing.

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